# Not quite Fermat's Last Theorem

Prove that the equation $n^a + n^b = n^c$, with $a,b,c,n$ positive integers, has infinite solutions if $n=2$, and no solution if $n\ge3$.

• (I know the answer, and I think it is nice enough to be worth posting)
– mau
Aug 2, 2010 at 13:07
• AFAIK, the theorem was first stated by <a href="en.wikipedia.org/wiki/Egbert_B._Gebstadter">Egbert B. Gebstadter</a>.
– mau
Aug 3, 2010 at 7:15

So this is fermats last theorem upside down? It occurs to me if we have two binary numbers we may add them to get another power of two,

   1000000
1000000
+ --------
10000000


but if we had two numbers in base 3, say

  1000000
1000000
+ -------
2000000


we would not have so much luck.

• Wow, this is an easy and elegant solution for an apparently difficult problem. Aug 2, 2010 at 14:54
• This is awesome! Aug 2, 2010 at 17:14
• Word to the wise: any Diophantine equation in which you can make all the terms only divisible by the same primes is easy. For example, it is very easy to solve variants of FLT of the form x^a + y^b = z^c provided a, b, c don't share too many common factors because one can just take x, y, z to be powers of the same integer. Aug 2, 2010 at 17:29
• @Bill: Why do you think this is not a proof? It's saying that if we add two numbers of the form 10… in base n > 2, we get either 20… if they're equal or 10…10… otherwise, and in neither case do we get a number of the form 10…. Seems a sufficiently detailed proof to me. Aug 2, 2010 at 21:02
• muad could have written: for the case n>=3, if we write n^a, n^b and n^c in base n, they will be of the form 1000..000, with the number of 0 in their expansion be equal to a, b, and c respectively. Now, the expression of the sum n^a + n^b written in base n is either is a number of the form 1000...0001000...000 if a != b or 2000...000 if a=b; in neither case we have a number of the form 1000....000. If n=2 and a=b, the sum becomes 1000...000 with a+1 zeroes, so this is a solution. // The procedure is the same. Or do you want a proof like in Principia Mathematica?
– mau
Aug 5, 2010 at 13:48

Wlog $\,a \le b$. Dividing by $n^a$ yields $\,1 + n^{b-a} = n^{c-a}$ $\Rightarrow$ $b=a\$ (else $\,n\mid1)\,$ $\Rightarrow$ $\, n = 2,\, c = a\!+\!1$.

• I am assuming the word Wlog stands for without loss of generality here. Feb 25, 2012 at 19:04
• Yes, that's what it means. Jan 29, 2014 at 15:37

If $n=2$ we can take $a=k, b=k, c=k+1$ for any $k \in \mathbb{N}$.

Let $n \ge 3$. We can assume that $a, b, c \ge 0$ because if not we could multiply left and right side by $n^k$ to make them positive.

Now it's clear that $c \ge a$ and $c \ge b$. Then we have $n^a | n^c$, hence $n^a | n^a + n^b$ and $a \le b$. In the same way $b \le a$. So $a = b$. Hence $2n^a = n^c$ and $n=2$.

• Man I hope they get this formula thing figured out soon. This answer looks like complete gibberish to me (I assume it's because all the the relevant formulae are just invisible). Aug 2, 2010 at 14:54

Assuming $b>a$:

$$n^b<n^a+n^b<n^{b+1}$$

• The $<n^{b+1}$ needs to be a $lt n^{b+1}$ for $a=1$, and I cannot see why it works for $n=2$ from this Aug 2, 2010 at 14:29
• @Tobias For n=2 it works only if a<b -- and for n=2 one indeed has infinitely many solutions (of the form a=b=c-1). (And I couldn't understand the first part of your comment.) Aug 2, 2010 at 14:36
• sorry, I was on a wrong track there, please ignore that comment Aug 3, 2010 at 7:29